# An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of 18  and the triangle has an area of 64 . What are the lengths of sides A and B?

Oct 16, 2016

$11.47$

#### Explanation:

We can use the Heron's formula for area of a Triangle

color(blue)("Area"=sqrt(s(s-a)(s-b)(s-c))

Where,

color(orange)("a, b and c are the sides"

color(orange)("s = semi-perimeter"=(a+b+c)/2

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Let's assign the values

color(purple)(a=x

color(purple)(b=x

color(purple)(c=18

color(purple)(s=(x+x+18)/2=(2x+18)/2

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Let's start to solve

The whole equation starts here

rArrcolor(violet)(sqrt((2x+18)/2((2x+18)/2-x)((2x+18)/2-x)((2x-18)/2-18))=64

The whole equation equals $64$ as, the area is $64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(\frac{2 x + 18 - 2 x}{2}\right) \left(\frac{2 x + 18 - 2 x}{2}\right) \left(\frac{2 x + 18 - 36}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(\frac{\textcolor{red}{2 x} + 18 - \textcolor{red}{2 x}}{2}\right) \left(\frac{\textcolor{red}{2 x} + 18 - \textcolor{red}{2 x}}{2}\right) \left(\frac{2 x - 18 - 36}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(\frac{\textcolor{red}{2 x - 2 x} + 18}{2}\right) \left(\frac{\textcolor{red}{2 x - 2 x} + 18}{2}\right) \left(\frac{2 x - 18 - 36}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(\frac{\textcolor{red}{18}}{2}\right) \left(\frac{\textcolor{red}{18}}{2}\right) \left(\frac{2 x - 18 - 36}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(\textcolor{red}{9}\right) \left(\textcolor{red}{9}\right) \left(\frac{2 x + 18 - 36}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(\textcolor{red}{81}\right) \left(\frac{2 x + 18 - 36}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(81\right) \left(\frac{2 x + \textcolor{\pi n k}{18 - 36}}{2}\right)} = 64$

$\rightarrow \sqrt{\frac{2 x + 18}{2} \left(81\right) \left(\frac{2 x \textcolor{\pi n k}{- 18}}{2}\right)} = 64$

Square both sides to take out the radical sign

$\rightarrow {\left(\sqrt{\frac{2 x + 18}{2} \left(81\right) \left(\frac{2 x - 18}{2}\right)}\right)}^{2} = {64}^{2}$

$\rightarrow \left(\frac{2 x + 18}{2}\right) \left(81\right) \left(\frac{2 x - 18}{2}\right) = \textcolor{red}{64 \cdot 64}$

$\rightarrow \left(\frac{2 x + 18}{2}\right) \left(81\right) \left(\frac{2 x - 18}{2}\right) = \textcolor{red}{4096}$

$\rightarrow \textcolor{\in \mathrm{di} g o}{\left(\frac{2 x + 18}{2}\right) \left(\frac{2 x - 18}{2}\right)} \left(81\right) = \textcolor{red}{4096}$

Take out the common factor $2$

$\rightarrow \textcolor{\in \mathrm{di} g o}{\left(\frac{2 \left(x + 9\right)}{2}\right) \left(\frac{2 \left(x - 9\right)}{2}\right)} \left(81\right) = \textcolor{red}{4096}$

$\rightarrow \textcolor{\in \mathrm{di} g o}{\left(\frac{\cancel{2} \left(x + 9\right)}{\cancel{2}}\right) \left(\frac{\cancel{2} \left(x - 9\right)}{\cancel{2}}\right)} \left(81\right) = 4096$

$\rightarrow \textcolor{\in \mathrm{di} g o}{\left(x + 9\right) \left(x - 9\right)} \left(81\right) = 4096$

Use the indentity color(brown)((a+b)(a-b)=a^2-b^2

$\rightarrow \textcolor{\in \mathrm{di} g o}{\left({x}^{2} - {9}^{2}\right)} \left(81\right) = 4096$

$\rightarrow \left({x}^{2} - 81\right) \left(81\right) = 4096$

$\rightarrow 81 {x}^{2} - 81 \cdot 81 = 4096$

$\rightarrow 81 {x}^{2} - 6561 = 4096$

$\rightarrow 81 {x}^{2} = 4096 + 6561$

$\rightarrow 81 {x}^{2} = 10657$

$\rightarrow {x}^{2} = \frac{10567}{81}$

$\rightarrow {x}^{2} = 131.56$

Take the square root of both sides

$\rightarrow \sqrt{{x}^{2}} = \sqrt{131.56}$

color(green)(rArrx=11.47